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3 years ago

Mystery number: The number is less than 190000 and greater than 180000.

1 answer:

6 0

The answer to it is
180, 000 <n> 190, 000
180,000 + 2000 = n = 5000-187, 000
182,000 = n = 182, 000
Or,
180, 000 <n> 190,000
n + 5000= 187, 000                           n – 2000 = 180, 000
n= 187, 000- 5000             n = 180, 000 + 2000
n= 182, 000                                       n= 182, 000

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How to solve this equation?

Novay_Z [31]

Answer:

x = √(28), or x = 5.292

Step-by-step explanation:

First you distribute the square to the values inside of the parentheses. so it ends up looking like this

5(x^2 - 25) - 9 = 6

add 9 to both sides

so its 5(x^2 - 25) = 15

Then distribute the multiplication of 5 to the contents within the parentheses

so it would be 5x^2 - 125 = 15

add 125 to both sides

you get 5x^2 = 140

divide by 5 on both sides

you get x^2=28

then, take the square root of both sides to reverse the square

√(x^2)=√(28)

and in the end you get x=5.292

but √(28) will probably be fine if your teacher doesn't want u to solve for that kind of stuff.

6 0

3 years ago

Three of the following statements are true. Which one is NOT true? <br> |-12| > 1

san4es73 [151]

|-12| is greater than 1

The absolute value brackets make the -12 positive, 12>1.

True

5 0

1 year ago

Read 2 more answers

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago

Answer this please!

Troyanec [42]

f(x) = 4x + 3

That's one-to-one, a linear function that will give a different f(x) for each different x.

For the inverse let's swap x and y and solve for y

x = 4y + 3

x - 3 = 4y

y = (x- 3)/4

That's the inverse,

Answer: (x- 3)/4, second choice

The domain and range of the inverse is all real numbers. It's two lines; I'll leave the graphing to you.

8 0

3 years ago

Can anyone help me with number 7 a b and c?

Sonbull [250]

Answer:

7a........they are similar because none of the sides are equal

6 0

3 years ago